20
$\begingroup$

Let $K$ be a finite extension of $\mathbf{Q}_p$. Let $F_d$ be the unramified extension of $\mathbf{Q}_p$ of degree $d$. I would like to know whether there exists some $d \geq 1$ and some $L \subset K \cdot F_d$ such that $L/\mathbf{Q}_p$ is totally ramified and $K \cdot F_d / L$ is unramified.

If $K/\mathbf{Q}_p$ is Galois, this can be done as follows. Take $d=[K:\mathbf{Q}_p]$. Write $Gal( F_d / \mathbf{Q}_p) = \langle \sigma \rangle$ and take any $\tilde{\sigma} \in Gal(K \cdot F_d / \mathbf{Q}_p)$ that lifts $\sigma$. Since $\tilde{\sigma}^d$ is the identity on both $K$ and $F_d$, it is the identity in $Gal(K \cdot F_d / \mathbf{Q}_p)$. We can then take $L = (K \cdot F_d)^{\langle \tilde{\sigma} \rangle}$.

Is the result still true if $K/\mathbf{Q}_p$ is not Galois?

$\endgroup$
13
$\begingroup$

This is not a complete answer, but perhaps it's a roadmap to a counterexample.

My strategy is to consider some non-Galois $K/\mathbf{Q}_p$ for which the result is true, and let's make some deductions about what $K$ must look like in this situation. Then let's find a $K$ that doesn't look like this.

Let $M/\mathbf{Q}_p$ denote the Galois closure of $K/\mathbf{Q}_p$. We're assuming the result is true, so let's choose $d$ and $L$ as in the question. Every field in the question is a subfield of $M.F_d$. Note also that the extension $K.F_d/L$ is unramified, and a composite of unramified extensions is unramified, so I am free to change $d$ to any multiple of $d$ and in particular I can assume that $d$ is a multiple of the degree of the maximal unramified subextension $M_0/\mathbf{Q}_p$ of $M/\mathbf{Q}_p$. In fact I really want to take $d=\infty$ and let $F_\infty$ denote the maximal unramified extension of $\mathbf{Q}_p$ in a fixed algebraic closure of $M$. So let's consider $M.F_\infty$, an infinite Galois extension of $\mathbf{Q}_p$, and let's see what the assertion that $L$ exists tells us about this situation.

We now have $F_\infty\subseteq L.F_\infty\subseteq K.F_\infty\subseteq M.F_\infty$. Recall that $M/\mathbf{Q}_p$ is Galois, with group $D$ say, and say $I$ is the inertia subgroup of $D$. Then $M.F_\infty/F_\infty$ is Galois with group $I$, and $M.F_\infty/\mathbf{Q}_p$ is Galois with group $=:D_M$, a semidirect product of $I$ and $\widehat{\mathbf{Z}}$ (with $I$ normal). The subfields $L.F_\infty$ and $K.F_\infty$ of the extension $M.F_\infty/F_\infty$ correspond to subgroups $I_L$ and $I_K$ of $I$, with $I_K\subseteq I_L$.

The tension in the situation is that $L/\mathbf{Q}_p$ is totally ramified -- we've not used this yet. Let's work in $Gal(M.F_\infty/\mathbf{Q}_p)$; recall this is a group with a finite subgroup $I$ and quotient $\widehat{\mathbf{Z}}$. The existence of $L$ from a Galois theory point of view tells us that there's a subgroup $D_L$ of this group with the property that $I_L$ is normal in $D_L$ (as $L.F_\infty/L$ is Galois) and such that the natural map from $Gal(L.F_\infty/L)$ to $Gal(F_\infty/\mathbf{Q}_p)$ is an isomorphism (here is where we use $L/\mathbf{Q}_p$ totally ramified).

In particular the normaliser of $I_L$ in $D_M$ must be quite big -- and I think that this is too much to ask. If $\sigma$ is a lift of Frobenius in $D_M$ then I think that $\sigma$ must normalise $I_L$. Note that $I_K$ and $I_L$ are both subgroups of $I$ and we know exactly how $\sigma$ acts on this, it's just the action coming from $D=Gal(M/\mathbf{Q_p})$.

So now let me imagine that I can choose $M/\mathbf{Q}_p$ such that $I=(\mathbf{Z}/p\mathbf{Z})^2$ with Frobenius acting as an automorphism of $I$ of order $p+1$ which cycles around all of the subgroups of order $p$. I think that I can explicitly build such an extension when $p=2$ by using Kummer theory on the unramified degree 3 extension of $\mathbf{Q}_2$.

Then if $I_K$ corresponds to a degree $p$ subgroup of $I$, we cannot have $I_L=I_K$ as the normaliser isn't big enough to contain $\sigma$, so we must have $I_L=I$. But this is bad because now $L$ corresponds to a subgroup of $D_M$ that contains $I$ and some lift of $\sigma$.

I have to see a student right now but what do you think? Sorry to be so sketchy!


Addition after seeing Doris' answer; So I think I now have an explicit counterexample.

There's a unique $A_4$ extension $M$ of $\mathbf{Q}_2$ apparently, with inertia the Sylow 2-subgroup and $f=3$. It's the splitting field of $x^4 - 2x^3 + 2x^2 + 2$. Take an element of order 2 in inertia and let $K$ be the corresponding degree 6 extension of $\mathbf{Q}_2$. We have $e=2$ for $K$ so if $L$ existed it would be a quadratic extension. Rather than count masses like Doris suggests, I am just going to do something far more moronic -- I will check using a computer that if $L/\mathbf{Q}_2$ is ramified and quadratic, then $ML/M$ is ramified and hence $KL/K$ must be too. There is probably a sensible way to do this but I just checked all 6 cases explicitly on a computer.

$\endgroup$
  • $\begingroup$ Nice counterexample! Nevertheless, I'll point out that my mass counting argument will show that there are counterexamples for pretty much all $e(K/\mathbb{Q}_p)>1$ and $f(K/\mathbb{Q}_p)>1$, and indeed it applies over any base field too. $\endgroup$ – Doris Feb 14 '17 at 14:36
  • $\begingroup$ @Doris: I'm sure you're right and I'm sure you're thinking about things in the right way. The reason I answered this question at all was because by chance I was thinking about a closely related problem on the way to work today. I wanted to find some explicit local field $K$ with the property that the absolute Frobenius automorphism on the residue field did not lift to $K$ (just to prove that such a thing can happen -- I am giving a talk on Monsky-Washnitzer cohomology tomorrow and one is always lifting Frobenius; I wanted to show that this was an assumption). $\endgroup$ – Kevin Buzzard Feb 14 '17 at 14:38
10
$\begingroup$

Note that the question is equivalent to the following: Given $K/\mathbb{Q}_p$, is there $L/\mathbb{Q}_p$ totally ramified so that $KL/K$ and $KL/L$ are unramified?

You note that it is true for $K/\mathbb{Q}_p$ Galois. It is also trivially true for $K/\mathbb{Q}_p$ totally ramified.

It is also true for $K/\mathbb{Q}_p$ tamely ramified: in this case you can write $K=\mathbb{Q}_p(\zeta,\sqrt[e]{\zeta^r p})$ where $\zeta$ is a primitive $p^f-1$th root of unity, so taking $L=\mathbb{Q}_p(\sqrt[e]{p})$ then $KL=K(\sqrt[e]{\zeta})=L(\sqrt[e]{\zeta})$ are unramified, as required.

I think that in general however, the answer is no. Fix the parameters $e=e(K/\mathbb{Q}_p)$ and $f=f(K/\mathbb{Q}_p)$, then we are looking for $L/\mathbb{Q}_p$ totally ramified of degree $e$ so that $KL/K$ and $KL/L$ are unramified. Assume $f>1$ and $e>1$ (because we know $f=1$ or $e=1$ works), then by Serre's mass formula (counting the number of totally ramified extensions of an extension of a p-adic field, which essentially is exponential in the degree of the base field over $\mathbb{Q}_p$) there are loads of possible $K$ ($n_K$ say), but not many possible $L$ ($n_L$ say). Let $d=(KL:K)$, then $(KL:\mathbb{Q}_p)=def=(KL:L)e$ so $(KL:L)=df$ is determined by $d$, and $d \leq e$. Hence the number of possible $KL$ by counting the $L/\mathbb{Q}_p$ is $n_Le$, whereas the number of different $K$ to consider is at least $n_K$, and assuming $n_Le < n_K$ (which is probably true if you look up the statement of the mass formula) then there are fewer solutions than questions, and hence some questions have no solution.

$\endgroup$
  • $\begingroup$ I thought a bit about $p=2$ so let me feed that into your answer. There is a unique $A_4$ extension of $\mathbf{Q}_2$ with inertia the Sylow 2-subgroup, so $e=4$ and $f=3$. Corresponding to a cyclic subgroup of order 2 in $I$ is a degree 6 extension $K/\mathbf{Q}_2$ with $e=2$ and $f=3$. My answer is a sketch of why I believe this should be a counterexample. With your eyes we see that we are now asking if there is $L/\mathbf{Q}_2$ ramified quadratic such that $LK/L$ and $LK/K$ are unramified. This looks unlikely to me and we can just check on a computer :-) $\endgroup$ – Kevin Buzzard Feb 14 '17 at 13:53
  • 1
    $\begingroup$ I just did this -- I'll add the details to my answer. $\endgroup$ – Kevin Buzzard Feb 14 '17 at 14:25
3
$\begingroup$

I'm a bit late but here are a few remarks on Kevin's example :

  1. There is a unique $\mathfrak{A}_4$-extension of $\mathbf{Q}_2$ because there is a unique cyclic cubic extension $C$ (namely the unramified one), the group $G=\mathrm{Gal}(C|\mathbf{Q}_2)=\mathbf{Z}/3\mathbf{Z}$ has a unique irreducible degree-$2$ $\mathbf{F}_2$-representation $\rho$, and $\rho$ occurs with multiplicity $1$ in $C^\times/C^{\times2}$. If $G$ acts on $D\subset C^\times/C^{\times2}$ through $\rho$, then Kevin's $M$ is $M=C(\sqrt D)$.

  2. Kevin's $K$ is not galoisian over $\mathbf{Q}_2$ (its galoisian closure is $M$) and nor is any of its unramified extensions, so no such $L$ exists (because an unramified extension of a quadratic extension is always galoisian).

  3. You can find $\mathfrak{A}_4$-extensions of ramification index $4$ and residual degree $3$ over every local field $F$ with finite residue field of characteristic $2$ (although they might not be unique, and in fact there are infinitely many of them if $F$ has characteristic $2$), so the argument can be made to work over every such $F$.

  4. $\mathfrak{A}_4$-extensions are galoisian closures of primitive quartic extensions (as are $\mathfrak{S}_4$-extensions, and these are the only two possibilities). I'm confident that the same trick can be applied over local fields with finite residue field of characteristic $p$ (arbitrary prime) by working with primitive extensions of degree $p^2$ or perhaps $p^n$ for some $n$. How does one find such extensions ? See https://arxiv.org/abs/1608.04183.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.